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Q. $B$ is the magnetic field at the centre of a circular loop. The magnetic moment of the circular loop of area $A$ is found to be $N\frac{BA^{Q}}{\mu _{0} \sqrt{\pi }}$ . The value of $2\left(N + Q\right)$ is -------.

NTA AbhyasNTA Abhyas 2022

Solution:

Let r be the radius of the circular loop.
$\therefore A=\pi r^{2}$
$\Rightarrow r=\sqrt{\frac{A}{\pi }}$
Magnetic field at the centre of the loop is
$B=\frac{\mu _{0} I}{2 r}$ $\Rightarrow B$
$=\frac{\mu _{0} I}{2 \sqrt{\frac{A}{\pi }}}\Rightarrow I=\frac{2 B}{\mu _{0}}\sqrt{\frac{A}{\pi }}$ Now, magnetic moment of the loop is
$M=IA$
$=\frac{2 B}{\mu _{0}}\sqrt{\frac{A}{\pi }}\cdot A$
$=\frac{2 B}{\mu _{0}}\frac{A^{\frac{3}{2}}}{\sqrt{\pi }}$ $\Rightarrow N=2$ and $Q=1.5$
$\therefore 2\left(N + Q\right)=7$